We are tasked with differentiating $f(x) = \sqrt{x}$ using the first principles of differentiation, as given by the formula:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Here, $f(x) = \sqrt{x}$, so let’s substitute that into the limit expression:

$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$

To simplify this expression, we multiply the numerator and denominator by the conjugate of the numerator:

$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}$

This gives:

$f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h})^2 - (\sqrt{x})^2}{h(\sqrt{x+h} + \sqrt{x})}$

Simplifying the numerator:

$f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$

Cancel the $h$ in the numerator and denominator:

$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$

Now, as $h \to 0$, $\sqrt{x+h} \to \sqrt{x}$. So the limit becomes:

$f'(x) = \frac{1}{2\sqrt{x}}$

Thus, the derivative of $f(x) = \sqrt{x}$ from first principles is:

$f'(x) = \frac{1}{2\sqrt{x}}$

Would you like any further clarification or a deeper explanation of any steps?

Here are 5 related questions to explore:

1. How would you apply first principles to differentiate $f(x) = x^2$?
2. What is the geometric interpretation of the derivative obtained for $f(x) = \sqrt{x}$?
3. How do higher-order derivatives work in first principles for more complex functions?
4. Can first principles be used to differentiate trigonometric functions, like $\sin(x)$?
5. How does the concept of limits apply when calculating derivatives?

Tip: When simplifying expressions using first principles, multiplying by the conjugate is a common strategy to handle square roots effectively.